Prove that $\inf(r,s)=r,\;\;\forall\;\;r,s\in \Re\;\;\text{and}\;\;s> r$
I tried proving the above but got stuck. Here is what I've done!
Set $\text{S}=(r,s)$, then
$$(i.)\;\;r<x\;\;\forall \;\;x\in \text{S} ,$$
$$(ii.)\;\;\text{Let}\;\;\epsilon>0 \;\;\text{be given. We want to find } $$
$$s_{\epsilon}\in \text{S}\;\;:\;\;r\leq s_{\epsilon}<r+\epsilon.$$
We may choose
$$s_{\epsilon}=r+\frac{\epsilon}{2},\;\;\text{if}\;\;\epsilon\leq s-r,\;\;\text{or}$$
$$=\text{any fixed }\;x_{0}\in \text{S}, \text{otherwise}.$$
Please, I'm I on track? If yes, can anyone help me complete the proof? This choice of $s_{\epsilon}$ does not interest me either! I'll love if someone can give a better choice of $s_{\epsilon}$. Various kinds of proofs are welcome!
We call $inf(r,s)=i$ if $i\in(r,s)$ then $\exists i' \in(r,i)$ .But then $i'<i$ which is not true. If $i\in (-\infty,r)$ then r does not exist in $(r,s)$. But we can find a greater bound $i''\in(i,r)$ also a contradiction.